Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{x^2 - 5x + 6}{x - 8} \div \dfrac{-7x + 21}{-2x + 16} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{x^2 - 5x + 6}{x - 8} \times \dfrac{-2x + 16}{-7x + 21} $ First factor the quadratic. $k = \dfrac{(x - 3)(x - 2)}{x - 8} \times \dfrac{-2x + 16}{-7x + 21} $ Then factor out any other terms. $k = \dfrac{(x - 3)(x - 2)}{x - 8} \times \dfrac{-2(x - 8)}{-7(x - 3)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (x - 3)(x - 2) \times -2(x - 8) } { (x - 8) \times -7(x - 3) } $ $k = \dfrac{ -2(x - 3)(x - 2)(x - 8)}{ -7(x - 8)(x - 3)} $ Notice that $(x - 8)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -2\cancel{(x - 3)}(x - 2)(x - 8)}{ -7(x - 8)\cancel{(x - 3)}} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $k = \dfrac{ -2\cancel{(x - 3)}(x - 2)\cancel{(x - 8)}}{ -7\cancel{(x - 8)}\cancel{(x - 3)}} $ We are dividing by $x - 8$ , so $x - 8 \neq 0$ Therefore, $x \neq 8$ $k = \dfrac{-2(x - 2)}{-7} $ $k = \dfrac{2(x - 2)}{7} ; \space x \neq 3 ; \space x \neq 8 $